AAMC MCAT Practice Exam 2
Multiple Choices Questions
with Latest 2025-2026 Correct
Answers GRADED A+
C/P: What expression gives the amount of light energy (in J per
photon) that is converted to other forms between the
fluorescence excitation and emission events?
"intensity of fluorescence emission at 440 nm excitation at 360
nm) was monitored for 20 minutes"
A) (6.62 × 10-34) × (3.0 × 108)
B) (6.62 × 10-34) × (3.0 × 108) × (360 × 10-9)
C) (6.62 × 10-34) × (3.0 × 108) × [1 / (360 × 10-9) - 1 / (440 ×
10-9)]
D) (6.62 × 10-34) × (3.0 × 108) / (440 × 10-9)
- ANS - C) (6.62 × 10-34) × (3.0 × 108) × [1 / (360 × 10-9) - 1 /
(440 × 10-9)]
The answer to this question is C because the equation of
interest is E = hf = hc/λ, where h = 6.62 × 10 −34 J ∙ s and c =
,3 × 10 8 m/s. Excitation occurs at λe = 360 nm, but
fluorescence is observed at λf = 440 nm. This implies that an
energy of E = (6.62 × 10 −34) × (3 × 10 8) × [1 / (360 × 10 −9) −
1 / (440 × 10 −9)] J per photon is converted to other forms
between the excitation and fluorescence events.
C/P: Compared to the concentration of the proteasome, the
concentration of the substrate is larger by what factor?
"purified rabbit proteasome (2 nM) was incubated in the
presence of porphyrin...the reaction was initiated by addition
of the peptide (100 uM)"
A) 5 × 101
B) 5 × 102
C) 5 × 103
D) 5 × 104
- ANS - D) 5 × 104
The answer to this question is D. The proteasome was present
at a concentration of 2 × 10-9 M, while the substrate was
present at 100 × 10-6 M. The ratio of these two numbers is 5 ×
104.
,sp2 hybridized
- ANS - possess exactly one doubly bonded atom
C/P: The concentration of enzyme for each experiment was 5.0
μM. What is kcat for the reaction at pH 4.5 with NO chloride
added when Compound 3 is the substrate?
Rate of reaction = 125 nM/s
A) 2.5 × 10-2 s-1
B) 1.3 × 102 s-1
C) 5.3 × 103 s-1
D) 7.0 × 105 s-1
- ANS - A) 2.5 × 10-2 s-1
The answer to this question is A. The fact that the rate of
product formation did not vary over time for the first 5
minutes implies that the enzyme was saturated with substrate.
Under these conditions, kcat = Vmax/[E] = (125 nM/s)/5.0 μM =
2.5 × 10-2 s-1.
kcat, Vmax, [E]
- ANS - kcat = Vmax/[E]
, C/P: Absorption of ultraviolet light by organic molecules
always results in what process?
A) Bond breaking
B) Excitation of bound electrons
C) Vibration of atoms in polar bonds
D) Ejection of bound electrons
- ANS - B) Excitation of bound electrons
The answer to this question is B. The absorption of ultraviolet
light by organic molecules always results in electronic
excitation. Bond breaking can subsequently result, as can
ionization or bond vibration, but none of these processes are
guaranteed to result from the absorption of ultraviolet light.
C/P: Four organic compounds: 2-butanone, n-pentane,
propanoic acid, and n-butanol, present as a mixture, are
separated by column chromatography using silica gel with
benzene as the eluent. What is the expected order of elution of
these four organic compounds from first to last?
A) n-Pentane → 2-butanone → n-butanol → propanoic acid
B) n-Pentane → n-butanol → 2-butanone → propanoic acid
C) Propanoic acid → n-butanol → 2-butanone → n-pentane
Multiple Choices Questions
with Latest 2025-2026 Correct
Answers GRADED A+
C/P: What expression gives the amount of light energy (in J per
photon) that is converted to other forms between the
fluorescence excitation and emission events?
"intensity of fluorescence emission at 440 nm excitation at 360
nm) was monitored for 20 minutes"
A) (6.62 × 10-34) × (3.0 × 108)
B) (6.62 × 10-34) × (3.0 × 108) × (360 × 10-9)
C) (6.62 × 10-34) × (3.0 × 108) × [1 / (360 × 10-9) - 1 / (440 ×
10-9)]
D) (6.62 × 10-34) × (3.0 × 108) / (440 × 10-9)
- ANS - C) (6.62 × 10-34) × (3.0 × 108) × [1 / (360 × 10-9) - 1 /
(440 × 10-9)]
The answer to this question is C because the equation of
interest is E = hf = hc/λ, where h = 6.62 × 10 −34 J ∙ s and c =
,3 × 10 8 m/s. Excitation occurs at λe = 360 nm, but
fluorescence is observed at λf = 440 nm. This implies that an
energy of E = (6.62 × 10 −34) × (3 × 10 8) × [1 / (360 × 10 −9) −
1 / (440 × 10 −9)] J per photon is converted to other forms
between the excitation and fluorescence events.
C/P: Compared to the concentration of the proteasome, the
concentration of the substrate is larger by what factor?
"purified rabbit proteasome (2 nM) was incubated in the
presence of porphyrin...the reaction was initiated by addition
of the peptide (100 uM)"
A) 5 × 101
B) 5 × 102
C) 5 × 103
D) 5 × 104
- ANS - D) 5 × 104
The answer to this question is D. The proteasome was present
at a concentration of 2 × 10-9 M, while the substrate was
present at 100 × 10-6 M. The ratio of these two numbers is 5 ×
104.
,sp2 hybridized
- ANS - possess exactly one doubly bonded atom
C/P: The concentration of enzyme for each experiment was 5.0
μM. What is kcat for the reaction at pH 4.5 with NO chloride
added when Compound 3 is the substrate?
Rate of reaction = 125 nM/s
A) 2.5 × 10-2 s-1
B) 1.3 × 102 s-1
C) 5.3 × 103 s-1
D) 7.0 × 105 s-1
- ANS - A) 2.5 × 10-2 s-1
The answer to this question is A. The fact that the rate of
product formation did not vary over time for the first 5
minutes implies that the enzyme was saturated with substrate.
Under these conditions, kcat = Vmax/[E] = (125 nM/s)/5.0 μM =
2.5 × 10-2 s-1.
kcat, Vmax, [E]
- ANS - kcat = Vmax/[E]
, C/P: Absorption of ultraviolet light by organic molecules
always results in what process?
A) Bond breaking
B) Excitation of bound electrons
C) Vibration of atoms in polar bonds
D) Ejection of bound electrons
- ANS - B) Excitation of bound electrons
The answer to this question is B. The absorption of ultraviolet
light by organic molecules always results in electronic
excitation. Bond breaking can subsequently result, as can
ionization or bond vibration, but none of these processes are
guaranteed to result from the absorption of ultraviolet light.
C/P: Four organic compounds: 2-butanone, n-pentane,
propanoic acid, and n-butanol, present as a mixture, are
separated by column chromatography using silica gel with
benzene as the eluent. What is the expected order of elution of
these four organic compounds from first to last?
A) n-Pentane → 2-butanone → n-butanol → propanoic acid
B) n-Pentane → n-butanol → 2-butanone → propanoic acid
C) Propanoic acid → n-butanol → 2-butanone → n-pentane
